Electric Shower tripping MCB

Page 1 of 3  
Hi all.
My brother is having a problem with his electric shower. You know the sort, the instantaneous ones which heat the mains cold water as it passes through. It keeps tripping the MCB. There is nothing else on the circuit, which is also RCD protected. It hasn't always happened, but it is now occuring very frequently. The MCB is rated at 40amps, and the shower is a 9.5 kilowatt unit (at 240volts).
Any ideas on what could cause the MCB to trip before the RCD, given that the shower shouldn't be pulling more than 40amps?
I guess the simple answer is that somehow the shower *is* pulling more than 40amps, or the MCB is naff and trips too easily.
Cheers Simon
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
Your MCB is an overcurrent protection device. It is tripping because it is undersised. For a 9.5 kW shower fit a 45 Amp MCB.

Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
On Sat, 17 Jan 2004, smb wrote:

In theory a 9.5kW shower will draw 39.6 Amps at 240 volts, however the supply voltage is allowed to go as high as 253 volts (230volts +- 10% IIRC) in which case it will draw significantly more. The MCB may also be on the sensitive side.
You should fit a 45 Amp MCB after checking that the cable to the shower, incoming supply and consumer unit are all properly sized.
--
Alistair Riddell - BOFH
Microsoft - because god hates us
  Click to see the full signature.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
You are a little wrong here, the supply voltage is allowed to fluctuate between +10% and -6%. However, should your voltage be higher than the 230 Volts, the less current your load will draw. This is a simple application of ohms law. The higher the voltage the less current the load will draw. That aside, as I said earlier - your MCB is too small. 45 Amps is the correct size for this shower.

Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

current
??? Since when did a shower have negative resistance!!!!!!
Volts=AMPS*RESISTANCE
Power=Volts*amps
If the resistance remains approximatly constant such as in the case of a shower if the voltage goes up the current goes up P=(V*V)/R
In the very very special case of a swich mode powersupply the output power and the input power is aprox constant regardless of supply voltage so ONLY in this special case the volts go up and the amps go down.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
You are wrong, P=V*I according to ohms law. Therefore, I= P/V Power remains constant @ 9.5 kW Therfore Amps = 43.2 @ 220 Volts Amps = 41.3 @ 230 Volts Amps = 39.6 @ 240 Volts Amps = 38 @ 250 Volts And just for the heck of it, Amps = 86.4 @ 110 Volts.
Do you see the reducing current with increasing voltage? Why do you think power is transmitted at such high voltages? Well the two main reasons are: High voltage, low current. Smaller cable size, cheaper. And secondly, voltage drop over long distances.

less
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
On Sat, 17 Jan 2004, ripper wrote:

That's where you are wrong, power does not remain constant. There are various means for modulating power to electric heating elements but electric showers do not generally incorporate anything more sophisticated than the ability to switch either or both of a pair of heating elements on.
Many heating appliances these days have rating plates showing the current consumption / power output for different input voltage e.g. 230v and 240v. In all cases you will find the power output (and current) higher for a higher input voltage.
--
Alistair Riddell - BOFH
Microsoft - because god hates us
  Click to see the full signature.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
wrote:

That part is true.
No, it doesn't. This is your fundamental error. The resistance of the shower element is approximately constant during operation. Increased voltage means increased current through the element, thus increased power. The power rating on the element is purely nominal.

No.
No argument with that....but it's a different topic.
--
Bob Eager
rde at tavi.co.uk
  Click to see the full signature.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
I am only using the 9.5 kW as an example. If you want to be fussy, the resistance will also change with temperature. So go find yourself an ohms law calculator, type in 9500 Watts, 240 Volts and tell us what the current is. Then try it with 9500 Watts, 230 Volts and tell us what the current is. I = P/V Yes? Well it did when I was at school.

Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
wrote:

And me too. But THE POWER ISN'T CONSTANT. I know the resistance will change with temperature. But not a lot.
Your ideas are so ridiculous, you must be NICEIC registered!
--
Bob Eager
rde at tavi.co.uk
  Click to see the full signature.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
ripper wrote on Saturday (17/01/2004) :

You are working on the false assumption that the 9.5Kw will remain constant irrespective of a change in voltage. It does not remain constant, the only constant is the resistance, but even this is not absolutely constant. As temperature rises, the resistance will also rise very slightly.
--

Regards,
Harry (M1BYT) (Lap)
  Click to see the full signature.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
ripper wrote on Saturday (17/01/2004) :

Because as V increases, I decreases for an identical load.

But that has nothing at all to do with the present discussion.
--

Regards,
Harry (M1BYT) (Lap)
  Click to see the full signature.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
On Sat, 17 Jan 2004, ripper wrote:

Right enough

I'm afraid you have it back to front - ohm's law is V=IxR (or I=V/R) therefore for a given resisitive load the current will increase proprotionally with the voltage.
--
Alistair Riddell - BOFH
Microsoft - because god hates us
  Click to see the full signature.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
Alistair You are trying to find current from two known quantities, voltage and power. Power remains constant, voltage varies. P=V*I P00 Watts Voltage = 230 Therefore I = P/V which is 9500/230 = 41.3 Amps the shower will take at 9.5 kW. Using the same power with a voltage of 240 Volts, I = P/V which is 9500/240 = 39.6 Amps. Now at 250 Volts, I = 9500/250 = 38 Amps. And if you use 110 Volts, the amps now increase to 86.4 Amps

current
less
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
And just to put you all at ease, if you go here you will find a simple ohms law calculator. http://measurementsconverter.co.uk/ohmslaw.html

fluctuate
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
wrote:

Nothing wrong with Ohm's Law...I learned it many yaers ago.
BUT...your application is wrong because of your erroneous assumption that power remains constant no matter what the applied voltage.
--
Bob Eager
rde at tavi.co.uk
  Click to see the full signature.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
Bob see my earlier post, I was not using the 9.5 kW as an assumption, merely an example on using ohms law to calculate current at various voltages for a fixed power rating. Perhaps it might be worth pointing out that the resistance of the heating element will also change with temperature. So at what temperature do we calculate the current draw for a 9.5 kW shower?

ohms
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
On Sat, 17 Jan 2004, ripper wrote:

That is where you are going wrong, the 9.5kW is NOT fixed, it is the power output quoted by the manufacturer for a given supply voltage.
The equations you are quoting are correct, but the constant value is the resistance of the element, NOT the power output.

It will, but not by much. A shower heater element is not like a light bulb whose resistance increases hugely when hot. An incandescent light bulb filament operates at thousands of degrees C, whereas a water heater element cannot go over 100 degrees C or you would be generating steam rather than hot water. (hopefully a safety cutout would intervene long before that point was reached).
--
Alistair Riddell - BOFH
Microsoft - because god hates us
  Click to see the full signature.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
On Sat, 17 Jan 2004, ripper wrote:

If you put 6.06 ohms into your calculator, this being a guess at the resistance of the heating element of the shower and the only constant [1] value in the equation, and then input different voltage values, then you will see the current and power increasing with higher voltage.
[1] I know it's not completely constant but near enough.
--
Alistair Riddell - BOFH
Microsoft - because god hates us
  Click to see the full signature.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
Yes that is what happens when you put a constant resistance into the calculator, and the wattage of the shower goes up. But the resistance does not remain constant, it varies with temperature.

ohms
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

Site Timeline

Related Threads

    HomeOwnersHub.com is a website for homeowners and building and maintenance pros. It is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.